SOLUTION: In a change purse there are some nickels some dimes and some quarters. There are 100 coins altogether and the number of nickels are the same as the number of dimes and quarters com
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Question 177294: In a change purse there are some nickels some dimes and some quarters. There are 100 coins altogether and the number of nickels are the same as the number of dimes and quarters combined. the total value of thecoins is $12.30. how many of each coin do i have?
> it can be answered by any means
> i got n=nickels
d=dimes
q=quarters
> n+d+q=100
> n=d+q
> .5n +.10d +.25q =$12.30
Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
n + d + q = 100
.05*n + .1*d + .25*q = 12.30
n = d + q
so far so good.
substituting in the first equation, we get:
d + q + d + q = 100
combining like terms:
2d + 2q = 100
dividing both sides by 2:
d + q = 50
d = 50 - q
since n = d + q
and d = 50 - q
then n = 50 - q + q = 50
we have:
n = 50
d = 50 - q
substituting in second equation of:
.05*n + .1*d + .25*q = 12.30
we get:
.05*50 + .1*(50-q) + .25*q = 12.3
simplifying we get:
2.5 + 5 - .1*q + .25*q = 12.3
combining like terms we get:
7.5 + .15*q = 12.3
subtracting 7.5 from both sides we get:
.15*q = 12.3 - 7.5 = 4.8
dividing both sides by .15 we get:
q = 4.8/.15 = 32
since d = 50 - q, this means that d = 50 - 32 = 18
we have:
n = 50
d = 18
q = 32
n + q + d = 50 + 18 + 32 = 100
.05*n + .1*d + .25*q = .05*50 + .1*18 + .25*32 = 2.5 + 1.8 + 8 = 12.3
both equations are satisfied, so the number of each coins is:
50 nickels
18 dimes
32 quarters
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